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Butoxors [25]
2 years ago
5

Pls help me answer this | | V

Mathematics
1 answer:
svp [43]2 years ago
7 0

Answer:

x=-6

Step-by-step explanation:

7x-12-2x=18

+12 both sides

7x-2x=30

7x-2x=-5x

-5x=30

divided both sides by -5

x=-6

hope this helps

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\sin w=\dfrac{1}{7}\\\\
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6 0
3 years ago
Harmony earns a \$42{,}000$42,000dollar sign, 42, comma, 000 salary in the first year of her career. Each year, she gets a 4\%4%
grandymaker [24]

Answer:

A=42000(1.04)^n

Step-by-step explanation:

This is a compound interest formula expressed as:

A=P(1+i)^n

Where:

  • n is time in years
  • i is the rate of interest
  • A is the accumulated amount after n years
  • P is the initial amount.

#We substitute the given values to determine amount after n years as follows:

A=P(1+i)^n\\\\=42000(1+0.04)^n\\\\=42000(1.04)^n

Hence, the amount earned after n years is given by the expression A=42000(1.04)^n

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2 years ago
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goldenfox [79]

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Step-by-step explanation:

7 0
3 years ago
A pair of pants with a marked tag of $35.00 has a discount of 20%. How much is the discount?
melisa1 [442]
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6 0
3 years ago
Suppose the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 5 million dollars
GuDViN [60]

Answer:

Probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.

Step-by-step explanation:

We are given that the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 5 million dollars. Also, incomes for the industry are distributed normally.

<em>Let X = incomes for the industry</em>

So, X ~ N(\mu=95,\sigma^{2}=5^{2})

Now, the z score probability distribution is given by;

         Z = \frac{X-\mu}{\sigma} ~ N(0,1)

where, \mu = mean income of firms in the industry = 95 million dollars

            \sigma = standard deviation = 5 million dollars

So, probability that a randomly selected firm will earn less than 100 million dollars is given by = P(X < 100 million dollars)

    P(X < 100) = P( \frac{X-\mu}{\sigma} < \frac{100-95}{5} ) = P(Z < 1) = 0.8413   {using z table]

                                                     

Therefore, probability that a randomly selected firm will earn less than 100 million dollars is 0.8413.

5 0
2 years ago
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